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Figure 21-14 represents two charged particles on an axis. However, a third particle is placed at a point to get the equilibrium condition for the three particles.

The condition of equilibrium due to any two charges on a third particle can be achieved if the forces due to the two charges are equal in magnitude and opposite in direction. For a given set of equal polar charges, the third particle should be of opposite polarity and placed between the charges to get the force direction to be opposite. Now, the equilibrium condition can be considered unstable if the movement of one charge will affect the movements of the other charges, and the state cannot be reversed; if not, then it is a stable one.

Formula:

The magnitude of the electrostatic force due to the two charges is $\left|F\right|=\frac{k\left|q_1\right|\left|q_2\right|}{r^2}$. (i)

Let us consider the separation between the given charges to be$\mathrm{d}$and the point is placed between them at a distance$x$from the charge role="math" localid="1662964730529" $-3q$;as for equal polarities of charges, the forces due to the two charges cancel out if the point is between them with opposite polarities. Thus, consider that the charge is $+q$.

Then, the condition of equilibrium at this point can be given using equation (i) as follows:

$$\begin{gathered} F_{-3q}=F_{-q} \\ \frac{k|-3q||+q^{\text{'}}|}{x^2}=\frac{k|-q||+q^{\text{'}}|}{{\left(d-x\right)}^2} \\ \frac{d-x}{x}=\frac{1}{\sqrt{3}} \end{gathered}$$

As this condition is justified, thus, the assumptions are correct.

From the above calculations and assumptions found to be correct, it can be said that the third particle is positively charged.

The found condition of equilibrium is unstable because if $+q$is moved slightly to the left, the attraction due to $-3q$will be more comparatively to attraction due to$-q$and it will not come back to its original form. Hence, the equilibrium is an unstable one.